91Ó°ÊÓ

Allow both gases to start at the same temperature. $T_i$, the initial volumes be $V_{i1}$and $V_{i2}$and let their number of moles be $n$. Two components of them are kept under constant strain. $p_1$and $p_2$, where indices $1$and $2$The monoatomic and diatomic gases are referred to in all quantities. We have the following for the received heat under constant pressure:

$Q_{monoatomic}=\frac{5}{2}nR\left(T_{f1}-T_i\right)$

$Q_{diatomic}=\frac{7}{2}nR\left(T_{f2}-T_i\right)$

where $T_{f1}$is the final temperature of monoatomic gas and $T_{f2}$is the final temperature of the diatomic gas.

We can see from the ideal gas law that at first

$p_1{V}_{i1}=nR{T}_i,p_2{V}_{i2}=nR{T}_i$

and in the final moment of expansion

$p_1{}^2·2V_{i1}=nR{T}_{f1},p_2·2V_{i2}=nR{T}_{f2}$

Substituting $p_1{V}_{i1}$and $p_2{V}_{i2}$from the first two relations into the other two we obtain

$T_{f1}=2T_i,T_{f2}=2T_i$

Applying this to the received heat expressions

$Q_{monoatomic}=\frac{5}{2}nR{T}_i$

$Q_{diatomic}=\frac{7}{2}nR{T}_i$

$Q_{monoatomic}=\frac{5}{2}nR{T}_i$

We may determine the appropriate ratio by dividing these two relations.

The acceptable ratio is $\frac{Q_{diatomic}}{Q_{monoatomic}}=\frac{7}{5}=1.4$